A couple years ago, a friend and I, while staying at a beachside condo, were standing on the balcony staring at the flat, distant horizon, and we got to wondering, “I wonder how far we can see right now?” and “I wonder how much further we can see up here on the 9th floor than the people out there on the beach?” Unable to resist, I got out a pen and paper and started drawing circles and right triangles, and we quickly had our answer. I was pleased that I still had the trigonometry knowledge to figure out the answer to such a question.

So hold on tight, because here we go, on a trigonometric journey to the end of the Earth!
First of all, we have to define what we mean by “How far to the horizon?” The simplest way, and the most common answer to the question if you ask Google, is to calculate the distance from the observer’s eyes to the horizon. This is a simple pythagorean triangle calculation. But what I mean, and what makes the most sense to me is to rephrase the question to, “How far would I have to travel along the surface of the earth to get to the last point I can see on the horizon?” That question is slightly trickier to solve. So let’s get going…

Diagram

Obviously this diagram is not drawn to scale. Let’s examine the components of the diagram:

Calculation

We know from trigonometry that the cosine of an angle of a right triangle is equal to the length of the adjacent side over the length of the hypotenuse.

So if we isolate theta…

Trigonometry also tells us that the length of an arc is the angle, in radians, of the arc multiplied by the radius.

So the final equation for the length of d in terms of h and r is:

If we plot this equation, plugging in the radius of the Earth, and dividing by 1000 to scale the y-axis to kilometers while the x-axis remains in meters, we get:

If we go up to an altitude of 50 meters, we can see 25 kilometers out to the horizon. This does, in fact, turn out to be extremely close to the square root of r / r + h that the Pythagorean theorem gives us for the length of the red line in the diagram. Only at very large values of h do the plotted lines (the length of the red line vs. the length of the yellow arc) diverge much.

Data Points

Let’s look at some concrete data points:

Meters Altitude

Kilometers To Horizon

1

3.572

2

5.051

3

6.186

4

7.143

5

7.986

6

8.749

7

9.450

8

10.102

9

10.715

10

11.294

In a chart, these numbers look like:

This means that a 200-cm-tall adult can see 1.5 km further than a 100-cm-tall child.

English System

For the metrically ignorant Americans, we’ll give the same data in terms they can understand:

Feet Altitude

Miles To Horizon

3

2.12

6

3.00

9

3.68

12

4.24

15

4.75

18

5.20

21

5.61

24

6.00

27

6.37

30

6.71

And in a chart:

So a 6-foot-tall adult, standing on the beach, can see exactly 3 miles to the horizon. You can see why, when looking for land, it’s advantageous to have someone looking out from a crow’s nest. At only 24 feet up, the visible distance doubles to 6 miles!

Distance To Horizon Calculator

While the topic is deeply interesting to me, the means of arriving at its conclusion makes my head hurt (which I suppose just paraphrases the above comment). Is it safe to assume that this works only in a case where the land is relatively smooth (or covered with water)? What does an interceding mountain range do to your calculations?

I also have to wonder if you’ve read Patrick O’Brian’s Aubrey-Maturin novels. If not, I bet you would love them.

http://www.erik-rasmussen.com/ Erik R.

Yep, Jake. You were the friend in the story. It was one February at the Ocean Reef. Good memory!

Jake

This sounds familiar but I canÂ’t pinpoint it.

Uncle Neil

I prefer looking around outside on my property. With just my own eyes in the clearest conditions I can see the hills 58 miles, 93 kilometers, across a shorter section of Lake Superior. The big freighters are starting to run by through there this week although we still have more than a meter of nice clean white snow.

Brings up something that has bothered me for ages:
Is the horizon the imaginary line that would be visible looking out over a flat landscape (e.g. the sea at sea level) or is it the actual apparent average of the limit between earth and sky?

One can climb peaks (in Southern California for example,) where you can see the horizon made by the ocean to the west, and the very flat (and near sea-level) desert to the west, but looking northward the horizon made by distant mountain ranges appears nearly as flat, and seems just barely bulged upwards. However, when you travel towards, and get up close to those other mountains, of course, they tower above you and create jagged, and distinct profiles.
Although there is nothing out at the sea’s horizon that can give you a point of reference to then judge your visibility distance on the map, out in the desert there are roads and intersections that can serve to tell you where your horizon lies.

As I have experimented, back in Scouts, and it’s what I expected, the distance you can see out over the desert is less than the distance to the horizon made be the mountains to the north.

**But do they count as the horizon, or as ‘visible objects beyond the horizon’ such as the Atlas Range I can sometimes see, beyond the edge of the Mediterranean’s horizon visible from here in Málaga?

http://www.erik-rasmussen.com/ Erik R.

This post was assuming a perfectly smooth sphere, simplified to a circle. Obviously things are a little different when you’ve got jagged mountains jutting up. Actually, it might be interesting to do another calculation of how far away point A and B could be given their altitudes HA and HB above sea level. Some of the commenters over on Bad Astronomy had the same questions as you, Ray.

I take personal offense to this line in your otherwise informative treatise:

“For the metrically ignorant Americans”

for I am an ‘American’ with a perfectly good mathematical sense of metric computing and can do it ‘in my head’, as I imagine most ‘Americans’ who stumble on this blog can, as well.

By The Way: Your tantalizing calculator at the bottom of the article doesn’t work for me on Firefox 3.5 or Safari 3.2

Cheers,

Jt

http://www.erik-rasmussen.com/ Erik R.

JT, first of all, I’d like to thank you for reporting the bug in the calculator. My blog has been through some database migration lately, and the calculator was a casualty. It’s fixed now.

I strongly suspect that you are naively optimistic about our fellow countrymen’s knowledge of the metric system. If you gave the following test to random Americans on the street (not just 13-year-old science students) I’d be very surprised if more than 10% got them all correct.

1) Which is longer, a centimeter or a kilometer?

2) How many centimeters are there in a meter?

3) How much does a liter of water weigh in kilograms?

4) Which is bigger, a liter or a gallon?

Most Europeans wouldn’t get #4 of course, because they have no concept of what a US gallon is. And I suspect that only the scientifically minded might get #3. But every single European would know the more practical question of, “Is 100 kph too slow or too fast to drive on the motorway (interstate to Americans)?” Only Americans that had spent some time abroad would have any clue about that question.

Keep in mind that ignorant doesn’t mean stupid; it means lacking in knowledge.

Josh

JT, you’re trolling, right? If you read the post carefully, you’ll see another opportunity to take “personal offense” — the assumption that the readers are trigonometrically ignorant. Obviously, most people old enough and sophisticated enough to be reading something like this blog are aware that there is no need to take personal offense when something does not allude to them personally. If, as you say, you have a “perfectly good mathematical sense of metric computing” then the “metrically ignorant” label doesn’t apply to you. See, no need to feel offended. [My guess is that Erik bothered with the polemic conversion table for the convenience of people like his mother (and in the case that his mother is actually comfortable with the metric system, than with my mother, and people like her).]

If what you really wanted to say is that you are offended that anybody would dare comment on the average American’s lack of knowledge of the metric system — a system of measurement which has been an international standard since the early 1960’s — than you might have a bone to pick more with the Secretary of Education than with Erik or any other stater of the obvious.

In short, the internet is such that you will waste far too much of your life getting offended by off-the-cuff comments which don’t even apply to you. Spend some time in NYC, or any other major urban center, and you’ll quickly harden your skin to such inconsequential offenses.

“If we shadows have offended, think but this, and all is mended, that you have but slumber’d here while these visions did appear. And this weak and idle theme, no more yielding but a dream…”

http://letterstosg.com Lance

Of course JT is trolling. Why else pretend to take offense at the statement that Americans are “metrically ignorant”? Of course we are. And, on a whole, we speak Spanish worse than Spaniards and Columbians. The average American also spends less time in Russia than the average Russian. And our turnout in the latest Japanese national elections was abysmal!

I would likewise challenge JT’s seriousness if he were to take offense (on behalf of, say, Spaniards) at my statement that Spaniards are relatively ignorant of US customary units.

Look, when it’s just Myanmar and Liberia who are refusing to adopt the metric system, the phrase “metrically ignorant” might sting. But when the world’s last superpower wants to use another system of measure, then it’s hardly a niche system. And one would hope that U.S. citizens would be better at the U.S. system than at others. There’s nothing offensive about saying so.

Allan Davidson

It seems to me that the easiest way to calculate the distance of the horizon is the old sailor’s method, which is:

The Square Root of the Height of Eye (in feet) x 1.16 = the distance of the visible horizon in Nautical Miles.

The same formula would have to be applied to any object which the observer wished to see, and then added to the first distance above to obtain the theoretical maximum distance that the top of the object to be viewed would become visible.

http://marcosmolina.wordpress.com Marcos Molina

Hello Erik, I am writing you from Mallorca, Spain. I happened to find your website while I was making the last post in my blog. I found extremely interesting the calculator to find out the distance to the horizon. I decided to link this post to my post (today 26-1-2010), in the context of my recent article Â«Among the horizon: theory and practiceÂ» ( http://marcosmolina.wordpress.com/2010/01/26/mas-alla-del-horizonte-y-vii-teoria-y-practica ), about the tops of the mountains in the Iberian peninsula that can be seen from the roof of my dear island Mallorca. Interestingly, these observations allow us to see much farther than the actual horizon thanks to circumstancial atmospheric refraction. I hope it’s ok for you to link your blog. Otherwise please let me know and I’ll delete it.
Congratulations for such an amazing website.
Best regards from Palma de Mallorca.
Marcos.

http://www.erik-rasmussen.com/ Erik R.

Most excellent post, Marcos! To everyone reading this, please go check out his gorgeous graphics, whether you can read Spanish or not.

Antonio Burla

Hi Erik… A quick one: all those calculations assume that at least one of the sides is on a plain looking into a horizon… What happens if both sides are on varying heights? How does it affect the formula? E.g. I live on the 30th floor of a building (say 100 metres in elevation) and I want to see how far into the horizon I can see if the objects there are, say, 400m in elevation, 1,000m in elevation and etc.

Rob Gulley

Antonio,

I spent a few days trying to figure out your question with both sides of varying height. I think this might be the answer:

Using:
R1 = Radius of earth
P = Height of your “platform” from the surface
M = Height of the mast of the boat you are watching sail away

Theta = Arcsin ( R1 / (R1+M) ) + Arcsin ( R1 / (R1+P) ) – PI/2

Then subtract Theta from Pi/2 to get the “Theta traveled by the boat” so to speak, that is, the angle formed by you, the boat, and the center of the earth. Divide the “Theta traveled” by two-PI, a full circle, and multiply by whatever you are using for the circumference of the earth. That’s how far the boat will be once the top of its mast drops below your horizon.

Eh?

Ahmet Faruk YAZICI

I came across this blog while trying to figure out if there is an add-on for Google Earth to calculate how far is the horizon for desired point.
In concept, it’s a very simple calculation. Google Earth provides the elevation data for every square-meter of our land. When you use this data with the equation on the “Horizon” article @wikipedia, you got what you need.

Jhonny Cris

I am doing this math project all about this and so I had to do it for other planets too like Saturn, but I wasn’t sure how to graph the formulas. When I saw this website and read it I was sooo exited and I think that I will get an A+! Thank you to the guy who made this website!

hselrahc

I live in Hawaii at an elevation of about 400 feet. We have an ocean view, where the ocean is about 4 1/2 miles away. From this calculation, our view of the ocean goes out to the horizon 24 1/2 miles away. Cool.

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Carsten

It’s not the “english” system. It’s called the imperial system

http://erikras.com/ Erik R.

You’re correct that “imperial” is more correct, but the term “English system” is often used in the US to refer to these “non-metric” units. And the actual “English System” from before 1824 did contain the two units, feet and miles, that I referred to in this post at their current values. So while I was definitely not as correct as I could’ve been, I’m certain that no misunderstandings could have occurred.

Loh Phat

I know we Americans (the Northern variant but not Canukistanians) are rightly lambasted as willfully ignorant, but road signage in the UK is STILL POSTED IN MILES not kilometers.